Equation of a straight line passing through the point of intersection of `x - y + 1 = 0 and 3x + y - 5 = 0` are perpendicular to one of them is

Answer

To find the equation of a straight line passing through the point of intersection of the lines \(x - y + 1 = 0\) and \(3x + y - 5 = 0\) that is perpendicular to one of these lines, we can follow these steps:

### Step 1: Find the Point of Intersection
We need to solve the two equations simultaneously to find their point of intersection.

1. From the first equation \(x - y + 1 = 0\), we can express \(y\) in terms of \(x\):
\[ y = x + 1 \]

2. Substitute \(y\) in the second equation \(3x + y - 5 = 0\):
\[ 3x + (x + 1) - 5 = 0 \]
\[ 3x + x + 1 - 5 = 0 \implies 4x - 4 = 0 \implies 4x = 4 \implies x = 1 \]

3. Now substitute \(x = 1\) back into \(y = x + 1\):
\[ y = 1 + 1 = 2 \]

Thus, the point of intersection is \((1, 2)\).

### Step 2: Determine the Slopes
Next, we need to find the slopes of the given lines.

1. For the line \(x - y + 1 = 0\):
- Rearranging gives \(y = x + 1\), so the slope \(m_1 = 1\).

2. For the line \(3x + y - 5 = 0\):
- Rearranging gives \(y = -3x + 5\), so the slope \(m_2 = -3\).

### Step 3: Find the Slope of the Perpendicular Line
To find the slope of the line that is perpendicular to one of the lines, we use the property that the product of the slopes of two perpendicular lines is \(-1\).

1. If we choose the line \(x - y + 1 = 0\) (with slope \(m_1 = 1\)):
\[ m \cdot 1 = -1 \implies m = -1 \]

2. If we choose the line \(3x + y - 5 = 0\) (with slope \(m_2 = -3\)):
\[ m \cdot (-3) = -1 \implies m = \frac{1}{3} \]

### Step 4: Write the Equation of the Perpendicular Line
Now we can write the equation of the line passing through the point \((1, 2)\) with the slopes found.

1. For the line perpendicular to \(x - y + 1 = 0\) (slope = -1):
\[ y - 2 = -1(x - 1) \]
\[ y - 2 = -x + 1 \implies x + y = 3 \]

2. For the line perpendicular to \(3x + y - 5 = 0\) (slope = \(\frac{1}{3}\)):
\[ y - 2 = \frac{1}{3}(x - 1) \]
\[ 3(y - 2) = x - 1 \implies 3y - 6 = x - 1 \implies x - 3y + 5 = 0 \]

### Final Result
The equations of the lines that are perpendicular to one of the given lines and pass through the point of intersection are:
1. \(x + y = 3\)
2. \(x - 3y + 5 = 0\)